Question

The value of ${\int }_1^{\frac{1+\sqrt{5}}{2}}\frac{x^2+1}{x^4-x^2+1}\log (1+x-\frac{1}{x})dx$

• A. $\frac{\pi }{8}{\log }_{e}2$
• B. $\frac{\pi }{2}{\log }_{e}2$
• C. $-\frac{\pi }{2}{\log }_{e}2$
• D. noneof these

Answer 1

The correct option is A $\frac{\pi }{8}{\log }_{e}2$

Let $I={\int }_1^{\frac{1+\sqrt{5}}{2}}\frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}\log (1+x-\frac{1}{x})dx$
$={\int }_1^{\frac{1+\sqrt{5}}{2}}\frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+1}\log (1+x-\frac{1}{x})dx$
Put $x-\frac{1}{x}=t$
$\Rightarrow \left(1+\frac{1}{x^2}\right)dx=dt$
Also, when $x=1\Rightarrow t=0$
And when $x=\frac{\sqrt{5}+1}{2}\Rightarrow t=1$
$\therefore I={\int }_0^1\frac{\ln (1+t)dt}{1+t^2}$
Put $t=\tan \theta$
$\Rightarrow dt={\sec }^2\theta d\theta$
$\therefore I={\int }_0^{\frac{\pi }{4}}\ln (1+\tan \theta )d\theta$
$I={\int }_0^{\frac{\pi }{4}}\ln (1+\tan (\frac{\pi }{4}-\theta \left)\right)d\theta$
$={\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1-\tan \theta }{1+\tan \theta })d\theta$
$={\int }_0^{\frac{\pi }{4}}\ln (\frac{2}{1+\tan \theta })d\theta$
$={\int }_0^{\frac{\pi }{4}}\ln 2-\ln (1+\tan \theta )d\theta$
$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\ln 2d\theta -{\int }_0^{\frac{\pi }{4}}\ln (1+\tan \theta )d\theta$
$2I={\int }_0^{\frac{\pi }{4}}\ln 2d\theta$
$2I=\ln 2[{\theta ]}_0^{\frac{\pi }{4}}$
$I=\frac{\pi }{8}\log 2$